Question

According to recent estimates by the Insurance Corporation of
British Columbia (ICBC), 20 percent of all auto insurance claims in
BC are fraudulent. Suppose that this estimate is correct,

(a) If 10 auto insurance claims are taken at random, use the
binomial distribution to determine the probability that:

(i) at least 6 claims are fraudulent

(ii) at most 3 claims are fraudulent

(iii) anywhere between 5 and 8 (inclusive) claims are
fraudulent

(iv) at least 1 claim is fraudulent

(b) If 15 auto insurance are taken at random, what is the
probability that between 2 and 6 claims (inclusive) are fraudulent
(binomial distribution)? Also using the Poisson approximation to
the binomial.

Answer #1

here this is binomial with parameter n=10 and
p=0.2 |

a)

i)

P(X>=6)=1-P(X<=5)= |
1-∑_{x=0}^{5}
(_{10}C_{x})p^{x}(q)^{(10-x)}
= |
0.0064 |

ii)

P(X<=3)= |
∑_{x=0}^{3 }
(_{10}C_{x})p^{x}(1−p)^{(10-x) }
= |
0.8791 |

iii)

P(5<=X<=8)= |
∑_{x=5}^{8 }
(_{n}C_{x})p^{x}(1−p)^{(n-x) }
= |
0.0328 |

iv)

P(X>=1)=1-P(X=0)=1-(1-0.2)^{10} =1-0.1074 =0.8926

b)

here this is binomial with parameter n=15 and
p=0.2 |

P(2<=X<=6)= |
∑_{x=2}^{6 }
(_{n}C_{x})p^{x}(1−p)^{(n-x) }
= |
0.8148 |

for Poisson approximation ; parameter =np=15*0.2 =3

P(2<=X<=6)= |
∑_{x=2}^{6} {e^{-3}*3^{x}/x!}= |
0.7673 |

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